Definite Integration Question 383
Question: The value of the integral $ \int_0^{\log 5}{\frac{e^{x}\sqrt{e^{x}-1}}{e^{x}+3}}dx= $
Options:
A) $ 3+2\pi $
B) $ 4-\pi $
C) $ 2+\pi $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ e^{x}-1=t^{2}\Rightarrow e^{x}dx=2tdt $
Also as $ x=0 $ to $ \log 5,t=0 $ to 2 Therefore, $ \int_0^{\log 5}{\frac{e^{x}\sqrt{e^{x}-1}}{e^{x}+3}}dx=\int_0^{2}{\frac{2t^{2}}{t^{2}+4}dt} $
$ =2[ \int_0^{2}{1dt-4\int_0^{2}{\frac{dt}{t^{2}+4}}} ]=4-\pi $ .
BETA
